Maximum degree of irreducible representation of group is less than or equal to product of maximum degree of irreducible representation of subgroup and index of subgroup

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Statement

Suppose G is a finite group, H is a subgroup, and K is a field whose characteristic does not divide the order of G (we do not require K to be a splitting field, though the splitting field case is of particular interest).

Then, the Maximum degree of irreducible representation (?) of G over K is less than or equal to the product:

(maximum degree of irreducible representation of H over K) times (index of H in G, i.e., [G:H])

Facts used

  1. Orthogonal projection formula
  2. Frobenius reciprocity

Related facts

Proof

Proof in characteristic zero

Given: G is a finite group, H is a subgroup, and K is a field of characteristic zero. H has index d in G. m is the maximum of the degrees of irreducible representations of G over K.

To prove: H has an irreducible representation over K whose degree is at least m/d.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let \alpha be the character of an irreducible representation of G of degree m. m is the maximum of degrees of irreducible representations of G.
2 Let \beta be the character of an irreducible representation of H arising as a subrepresentation of \operatorname{Res}_H^G \alpha. H is a subgroup of the finite group G.
3 The inner product \langle \operatorname{Res}_H^G \alpha, \beta \rangle_H is a positive integer, because \beta is the character of a subrepresentation of the representation affording \operatorname{Res}_H^G\alpha. (Note: Over a splitting field, the inner product is precisely the multiplicity, but it may be bigger in general if \beta is not absolutely irreducible). We need K to have characteristic zero to make sense of positive integer.
4 \langle \operatorname{Res}_H^G \alpha, \beta \rangle_H = \langle \alpha, \operatorname{Ind}_H^G \beta \rangle Fact (2)
5 \langle \alpha, \operatorname{Ind}_H^G \beta \rangle is a positive integer, indicating that \alpha is an irreducible constituent of \operatorname{Ind}_H^G \beta. Steps (3), (4)
6 The degree of \alpha, which is m, is less than or equal to the degree of \operatorname{Ind}_H^G\beta Steps (1), (5)
7 The degree of \operatorname{Ind}_H^G\beta is d times the degree of beta H has index d in G. Definition of induced representation.
8 The degree of beta is at least m/d Steps (6), (7)